Question

graph the feasible region for the follow system of inequalities by drawing a polygon around the feasible region. click to set the corner points.

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Explanation:

Step1: Simplify the first inequality

Divide \(3y + 6x\geq27\) by \(3\) to get \(y+2x\geq9\), which can be rewritten as \(y\geq - 2x + 9\).

Step2: Rewrite the second inequality

From \(4y+x\geq9\), we get \(y\geq-\frac{1}{4}x+\frac{9}{4}\).

Step3: Analyze the third and fourth inequalities

The inequalities \(y + x\geq6\) (or \(y\geq - x + 6\)) and \(y + x\leq9\) (or \(y\leq - x+9\)) along with \(x\geq0\) and \(y\geq0\).

Step4: Find intersection points

• Intersection of \(y=-2x + 9\) and \(y=-x + 6\):

Set \(-2x + 9=-x + 6\), then \(x = 3\) and \(y=3\).

• Intersection of \(y=-2x + 9\) and \(y=-x + 9\):

Set \(-2x + 9=-x + 9\), then \(x = 0\) and \(y=9\).

• Intersection of \(y=-\frac{1}{4}x+\frac{9}{4}\) and \(y=-x + 6\):

Set \(-\frac{1}{4}x+\frac{9}{4}=-x + 6\), \(-x+9=-4x + 24\), \(3x=15\), \(x = 5\) and \(y=1\).

• Intersection of \(y=-\frac{1}{4}x+\frac{9}{4}\) and \(y=-x + 9\):

Set \(-\frac{1}{4}x+\frac{9}{4}=-x + 9\), \(-x + 9=-4x+36\), \(3x = 27\), \(x=9\) and \(y=0\).

• Intersection of \(y=-x + 6\) and \(x = 0\): \(y=6\) (point \((0,6)\)).
• Intersection of \(y=-x + 9\) and \(y = 0\): \(x=9\) (point \((9,0)\)).

Answer:

The corner - points of the feasible region (by checking the boundaries and their intersections with non - negativity constraints) are \((0,9)\), \((3,3)\), \((5,1)\), \((9,0)\)